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2 years ago

A snowmobile has an initial velocity of 3.0 m/s

2 answers:

7 0

My answer to the problem is as follows:

<span>1. Use the kinematic formula

Vf = Vi + a*t

for a, Vi = 3.0 m/s, a = 0.5 m/s/s, and t = 7.o s.

for b, Vf = 0, Vi = 3.0 m/s, and a = -0.60 m/s/s.

I hope my answer has come to your help. God bless and have a nice day ahead!
</span>

bazaltina [42]2 years ago

3 0

Answer:

Part a)

$v_f = 6.5 m/s$

Part b)

$t = 5 seconds$

Explanation:

Part a)

As we know that

$v_f = v_i + at$

here we know that

$v_i = 3 m/s$

$a = 0.5 m/s^2$

$t = 7.0 s$

so we will have

$v_f = 3 + (0.5)(7.0)$

$v_f = 6.5 m/s$

Part b)

if finally the snowmobile comes to rest

So here we can say that

$v_f = 0$

$v_i = 3.0 m/s$

$a = -0.60 m/s^2$

so now we have

$v_f = v_i + at$

$0 = 3.0 - (0.60)t$

$t = 5 s$

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The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

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Answer:

The spring constant = 104.82 N/m

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Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

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k = spring constant

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m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

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